# Plis solve it using step by step! Logarithms.

Solve for x: 9^x-3^x-8=0

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1. since 9^x = (3^2)^x = (#^x(^2
we can let 3^x = y , then
9^x - 3^x - 8 = 0 becomes
y^2 - y - 8 = 0
y = (1 ± √33)/2
y = (1+√33)/2 or y = (1-√33)/2

3^x = (1+√33)/2
take log of both sides, and use log rules
x = log[(1+√33)/2] /log3 = appr 1.106

or

3^x = (1-√33)/2 which has no solution, since we will not be able to take the log of a negatiave number.

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2. (3^2)^x-3^x-8=0,
(3^x)² -3^x -8 = 0,
3^x =y,
y² -y - 8=0,
y= (1±√33)/2.
Since 3^x>0, the negative root is nonsence,
=> y= (1+√33)/2 =3.37228,
3^x =3.37228,
x=log(3)3.37228= =ln3.37228/ln3≈1.21559/1.0986=1.1065

Test:
9^1.1065 – 3^1.1065 – 8 =
=11.372 – 3.372 – 8 = 0

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